91Ó°ÊÓ

1. The length of pedal arm is, $r=0.152m$.
2. The downward force is, $F=111N$
3. ${\mathrm{θ}}_1={30}^o$
4. ${\mathrm{θ}}_2={90}^o$
5. ${\mathrm{θ}}_3={180}^o$
   

Torqueis a turning action on a body about a rotation axis due to a force. If force is applied at a point, then total torque is the cross product of radial vector and force exerted on the body. The magnitude of torque is$\mathit{τ}\mathbf{=}\mathit{r}\mathit{F}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}$

$\begin{array}{rcl}\mathit{τ}& \mathbf{=}& \mathit{r}\mathbf{×}\mathit{F}\\ & \mathbf{=}& \mathit{r}\mathit{F}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\end{array}$

According to the formula, we can calculate the magnitude of torque for${\mathrm{θ}}_1={30}^o$

$$\begin{gathered} \mathrm{τ}=r×F \\ \mathrm{τ}=rF\sin \mathrm{θ}. \end{gathered}$$

Substitute all the value in the above equation.

$\begin{array}{rcl}\mathrm{τ}& =& 0.152m×111N×\sin 30\\ & =& 8.44N.m\\ \mathrm{τ}& ≈& 8.4N.m\end{array}$

Hence the torque is, 8.4N.m .

According to formula we can calculate magnitude of torque for ${\mathrm{θ}}_2={90}^o$

$$\begin{gathered} \mathrm{τ}=r×F \\ \mathrm{τ}=rF\sin \mathrm{θ} \end{gathered}$$

Substitute all the value in the above equation.

$\begin{array}{rcl}\mathrm{τ}& =& 0.152m×111N×\sin 90\\ & =& 16.90N.m\\ \mathrm{τ}& ≈& 17N.m\end{array}$

Hence the torque is, 17N.m .

According to formula we can calculate magnitude of torque for${\mathrm{θ}}_3={180}^o$

$$\begin{gathered} \mathrm{τ}=r×F \\ \mathrm{τ}=rF\sin \mathrm{θ} \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} \mathrm{τ}=0.152m×111N×\sin 180 \\ \mathrm{τ}=0N.m \end{gathered}$$

Hence the torque is, 0N.m.